Archive for January 2012

Do there exist powers of 2 whose first four digits are 2012?
Are there Fibonacci numbers whose first four digits are 2012?

If the answer is obvious to you (or you don’t care), you can stop reading.

The answer for both is:

Yes.

There are infinitely many such numbers.

In fact, the fraction of (powers of 2 / Fibonacci numbers) starting with 2012 is exactly

Similarly with any other prefix p (of any length) in place of 2012. Proof follows.

A number x starts with a prefix p if and only if for some k ≥ 0,

Thus a power of 2, say 2n, starts with p iff

for some

Taking logarithms to base 10 and simplifying, this is equivalent to

for some

This is saying that the fractional part of lies between the fractional parts of and For example, if , this means that the fractional part of lies between and .

Similarly, for Fibonacci numbers, as is (or should be) well-known, the nth Fibonacci number Fn is the closest integer to , where is the golden ratio. So starts with iff

Taking logarithms to base 10 and simplifying, while ignoring the annoying which becomes irrelevant in the limit (this line is not rigorous), this is equivalent to

which means that the fractional part of lies between the fractional parts of and . For , this means that the fractional part of lies between and .

In either case, we are trying to make the fractional part of , for some irrational number , lie in some interval. The relevant fact is this:Theorem 1: for any irrational number , the sequence (where denotes the fractional part of ) is dense in .
or, in other words,Theorem 1: For any irrational number , the sequence is dense modulo .
Proving this theorem is a good exercise.

This means that for any interval you want, you can always find some such that the fractional part of lies in your interval. In fact, because the sequence is dense, you can find an infinite sequence of such that the fractional parts of converge to the midpoint (say) of the desired interval. This proves the first two facts of the answer, and for the third we need a stronger theorem: